Problem: Complete the square to solve for $x$. $x^{2}-13x+40 = 0$
Explanation: Move the constant term to the right side of the equation. $x^2 - 13x = -40$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $-13$ , so half of it would be $-\dfrac{13}{2}$ , and squaring it gives us ${\dfrac{169}{4}}$ $x^2 - 13x { + \dfrac{169}{4}} = -40 { + \dfrac{169}{4}}$ We can now rewrite the left side of the equation as a squared term. $( x - \dfrac{13}{2} )^2 = \dfrac{9}{4}$ Take the square root of both sides. $x - \dfrac{13}{2} = \pm\dfrac{3}{2}$ Isolate $x$ to find the solution(s). $x = \dfrac{13}{2}\pm\dfrac{3}{2}$ The solutions are: $x = 8 \text{ or } x = 5$ We already found the completed square: $( x - \dfrac{13}{2} )^2 = \dfrac{9}{4}$